B Question regarding Heat Transfer in Carnot Engine

AI Thread Summary
A Carnot Engine is a theoretical model that differs from practical engines like the Stirling Engine. The heat transfer in a Carnot Engine occurs only during isothermal processes, leading to irreversible heat loss if the working material is not at the same temperature as the hot reservoir. The discussion raises a question about how the engine can transfer heat to the cold reservoir if all heat supplied is converted into work. It is clarified that the work done is the difference between the heat supplied and the heat rejected, not solely the heat supplied. The adiabatic process is essential for completing the cycle, effectively halting heat transfer, though achieving this in practice is challenging.
Harikesh_33
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***A Carnot Engine*** is a theoretical engine unlike a Sterling Engine which can be made practically.

Some of the drawbacks of Carnot Engine are,

1)The Heat Transfer occurs only during isothermal process(compression and expansion),this is because the working material (ie) gas or fuel used, if it's at a different temperature(technically the temperature of the working material is infinitesimally smaller)than the hot reservoir ,then some amount of heat get's used up to attain thermal equilibrium (ie) it's irreversibly lost .

Now my question is ,if ##Q_H## is the heat supplied by the #Heat# #Reservoir# ,then the heat is used to do some work of say ##dW=pdV## and if entire heat ##Q_H## gets used as Work, then how does it transfer ##Q_c## amount of heat to the cold reservoir?

And it seems so odd to say that once the Isothermal processes are over ,it expands /contracts adiabatically ,are we physically removing the sources here ? **Is the only use of Adiabatic process in ***"Imaginary Carnot Cycle"*** to produce a proper cycle**?
 
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Harikesh_33 said:
Now my question is ,if ##Q_H## is the heat supplied by the #Heat# #Reservoir# ,then the heat is used to do some work of say ##dW=pdV## and if entire heat ##Q_H## gets used as Work, then how does it transfer ##Q_c## amount of heat to the cold reservoir?
The amount of work is ##Q_H - Q_C##, not just ##Q_H##.

Harikesh_33 said:
are we physically removing the sources here ? **Is the only use of Adiabatic process in ***"Imaginary Carnot Cycle"*** to produce a proper cycle**?
Yes, you are shutting down the heat transfer process, Something similar to closing a valve to shut down a mass flow. Something very difficult to achieve physically, most likely impossible at high speed.
 
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